Question

我在课堂上有一个链接列表，如

7,5,3,1,2,4,6,8

我想得到像

这样的输出

1,3,5,7,2,4,6,8

当我尝试使用以下反向代码时：

class Linkedlist:
def __init__(self, L = None):
...
def partlyreverse(self):
    tail = self.head
    current_node = self.head.next_node
    tail.next_node = None
    while current_node.value % 2 == 1:
        next_current_node = current_node.next_node
        current_node.next_node = tail
        tail = current_node
        current_node = next_current_node

    self.head = tail
L = Linkedlist([7,5,3,1,2,4,6,8])
L.partlyreverse()

我只能

1,3,5,7

所以我怎么能这样做？

谢谢！

Answer 1

你基本上需要检查列表的开头是否包含奇数或偶数！基于此你需要逆转！

>>> s
(7, 3, 5, 1, 2, 4, 6, 8)
>>> sorted(filter(lambda x:(x%2!=0 and s[0]%2!=0) or (x%2==0 and s[0]%2==0),s)) + sorted(filter(lambda x:(x%2!=0 and s[-1]%2!=0) or (x%2==0 and s[-1]%2==0),s))
[1, 3, 5, 7, 2, 4, 6, 8]

Answer 2

你的问题不是100％明确：你是否试图将链表分成两半，开头是奇数整数，最后是偶数？您是否尝试对列表进行分区而不考虑值本身？

此外，必须有您未提交问题的代码...我已经最好地猜测了代码的外观。

下面的代码段会返回您想要的输出：

class Node:
    def __init__(self, value):
        self.value = value
        self.next_node = None

class Linkedlist:
    def __init__(self, L=None):
        self.head = Node(L[0])
        current = self.head
        for value in L[1:]:
            current.next_node = Node(value)
            current = current.next_node

    def partlyreverse(self):
        tail_start = self.head
        tail_end = self.head
        tail2_start = None
        tail2_end = None

        current_node = self.head.next_node

        while current_node != None:
            if current_node.value % 2 == 1:
                next_current_node = current_node.next_node
                current_node.next_node = tail_start
                tail_start = current_node
                current_node = next_current_node
            else:
                if not tail2_start:
                    tail2_start = current_node
                    tail2_end = tail2_start
                    current_node = current_node.next_node             
                else:
                    tail2_end.next_node = current_node
                    tail2_end = current_node
                    current_node = current_node.next_node 

        tail_end.next_node = tail2_start
        self.head = tail_start

L = Linkedlist([7,5,3,1,2,4,6,8])

L.partlyreverse()

def RecursivelyPrintNodes(node):
    if node != None:
        print node.value
        RecursivelyPrintNodes(node.next_node)

RecursivelyPrintNodes(L.head)

# 1,3,5,7,2,4,6,8

最后，我在你的代码中看到了你的目标，但是作为一个注释，你编写的代码将无法编译（你的类中存在缩进错误）。

python linkedlist部分反转

2 个答案: